bigg bio lab 2
Part I: Pre-Lab
Diffusion/Osmosis Pre-Lab
Introduction: Dialysis tubing allows molecules to diffuse through microscopic pores in the tubing. Molecules that are smaller than the pores can diffuse through the dialysis membrane along the concentration gradients. Molecules that are larger than the pore size are prevented from crossing the dialysis membrane.
Answer questions in complete sentences. For problems, show equations and work with units and appropriate significant figures.
Part IA:
In the following situations, assume the sucrose cannot diffuse through the dialysis membrane.
If a dialysis bag contains a 0.2 M solution of sucrose is placed in a beaker of distilled water, will the dialysis bag gain or lose mass? Explain why.
Hypothesis: It will lose mass
Explanation: The solution will move out in order to be balanced with the environment surrounding it
A dialysis bag has an initial mass of 26.3 g and a final mass of 30.2 g. Find the percent change in mass. Show your work!
Work: (mass initial – mass final)/(mass initial) x 100%= 14.83%
Part IB:
With the data below, calculate the percent change in mass, graph the data, determine where the percent change in mass crosses the X axis, determine the solute potential of the potato if the room temperature is 24oC, and finally determine the water potential of the potato.
Contents in Dialysis Tubing
Initial Mass in grams
Final Mass in grams
Percent Change in Mass
0.0 M
22.0
27.0 22.73%
0.2 M
24.6
26.4 7.32%
0.4 M
23.5
23.2 -1.28%
0.6 M
23.7
20.4 -13.93%
0.8 M
19.9
15.6 -21.61%
1.0 M
21.3
16.2 -23.93% Where the line crosses the X axis0.3M
Solute potential of the potato: 14.81 bars
Water potential of the potato: 14.81 bars
Work:
Solute potential: (ψS) = – iCRT = (-2)(.3)( 0.0831)(297)=14.81 bars
Water potential: ψ = ψP + ψS = = water potential= 0 + 14.81= 14.81
LAB PART 2
Part 1: What factors limit cell size?
1. What is surface area? Area of given surface
2. What is volume? Area inside a shape
3.
Diffusion/Osmosis Pre-Lab
Introduction: Dialysis tubing allows molecules to diffuse through microscopic pores in the tubing. Molecules that are smaller than the pores can diffuse through the dialysis membrane along the concentration gradients. Molecules that are larger than the pore size are prevented from crossing the dialysis membrane.
Answer questions in complete sentences. For problems, show equations and work with units and appropriate significant figures.
Part IA:
In the following situations, assume the sucrose cannot diffuse through the dialysis membrane.
If a dialysis bag contains a 0.2 M solution of sucrose is placed in a beaker of distilled water, will the dialysis bag gain or lose mass? Explain why.
Hypothesis: It will lose mass
Explanation: The solution will move out in order to be balanced with the environment surrounding it
A dialysis bag has an initial mass of 26.3 g and a final mass of 30.2 g. Find the percent change in mass. Show your work!
Work: (mass initial – mass final)/(mass initial) x 100%= 14.83%
Part IB:
With the data below, calculate the percent change in mass, graph the data, determine where the percent change in mass crosses the X axis, determine the solute potential of the potato if the room temperature is 24oC, and finally determine the water potential of the potato.
Contents in Dialysis Tubing
Initial Mass in grams
Final Mass in grams
Percent Change in Mass
0.0 M
22.0
27.0 22.73%
0.2 M
24.6
26.4 7.32%
0.4 M
23.5
23.2 -1.28%
0.6 M
23.7
20.4 -13.93%
0.8 M
19.9
15.6 -21.61%
1.0 M
21.3
16.2 -23.93% Where the line crosses the X axis0.3M
Solute potential of the potato: 14.81 bars
Water potential of the potato: 14.81 bars
Work:
Solute potential: (ψS) = – iCRT = (-2)(.3)( 0.0831)(297)=14.81 bars
Water potential: ψ = ψP + ψS = = water potential= 0 + 14.81= 14.81
LAB PART 2
Part 1: What factors limit cell size?
1. What is surface area? Area of given surface
2. What is volume? Area inside a shape
3.